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VladimirAG [237]
1 year ago
9

What is the only additional information needed to calculate the empirical formula of a compound if the masses of each element of

the compound are provided?
Chemistry
1 answer:
Natasha_Volkova [10]1 year ago
5 0

Mass percentages of the the elements in the compound is the only additional information needed to calculate the empirical formula of a compound if the masses of each element of the compound are provided.

<h3>What is Empirical Formula ? </h3>

The Empirical formula is the simplest whole number ratio of atoms present in given compound.

<h3>What is Molecular formula of Compound ?</h3>

Molecular Formula = Empirical Formula × n

n =  \frac{\text{Molar Mass}}{\text{Empirical Formula weight}}

Thus from the above conclusion we can say that Mass percentages of the the elements in the compound is the only additional information needed to calculate the empirical formula of a compound if the masses of each element of the compound are provided.

Learn more about Empirical Formula here: brainly.com/question/1603500

#SPJ4

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What approximate percentage of the Earth’s freshwater is groundwater?
grigory [225]

the answer is c.) 30%

4 0
3 years ago
Read 2 more answers
How many half-lives will pass by the time 1.56% of I-131 is present? B. Approximately how many days does that equal? *
serg [7]

Answer: Hmmmmm that's crazy....

There are a couple of equations one could use for this type of problem, but I find the following to be the easiest to use and to understand.

Fraction remaining (FR) = 0.5n

n = number of half lives that have elapsed

In this problem, we need to find n and are given the FR, which is 1.56% or 0.0156 (as a fraction).

0.0156 = 0.5n

log 0.0156 = n log 0.5

-1.81 = -0.301 n

n = 6.0 half lives have elapsed

Explanation:

Just wanted to help. Hopefully it's correct wouldn't want to waster your time ;)

6 0
2 years ago
The rate constant k for a certain reaction is measured at two different temperatures temperature 376.0 °c 4.8 x 108 280.0 °C 2.3
9966 [12]

Answer:

The activation energy for this reaction = 23 kJ/mol.

Explanation:

Using the expression,

\ln \dfrac{k_{1}}{k_{2}} =-\dfrac{E_{a}}{R} \left (\dfrac{1}{T_1}-\dfrac{1}{T_2} \right )

Where,

k_1\ is\ the\ rate\ constant\ at\ T_1

k_2\ is\ the\ rate\ constant\ at\ T_2

E_a is the activation energy

R is Gas constant having value = 8.314×10⁻³ kJ / K mol  

k_2=2.3\times 10^8

T_2=280\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (280 + 273.15) K = 553.15 K  

T_2=553.15\ K  

k_1=4.8\times 10^8  

T_1=376\ ^0C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (376 + 273.15) K = 649.15 K  

T_1=649.15\ K  

So,  

\left(\ln \left(\:\frac{4.8\times \:\:\:10^8}{2.3\times \:\:\:10^8}\right)\right)\:=-\frac{E_a}{8.314\times \:10^{-3}\ kJ/mol.K}\times \:\left(\frac{1}{649.15\ K}-\frac{1}{553.15\ K}\right)

E_a=-\frac{10^{-3}\times \:8.314\ln \left(\frac{10^8\times \:4.8}{10^8\times \:2.3}\right)}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=-\frac{\frac{8.314\ln \left(\frac{4.8}{2.3}\right)}{1000}}{-\frac{96}{359077.3225}}\ kJ/mol

E_a=22.87\ kJ/mol

<u>The activation energy for this reaction = 23 kJ/mol.</u>

6 0
2 years ago
Identify each of the following as a representative element or a transition element:
charle [14.2K]
The correct answer is B. Platinum is the transition element among the choices. The elements belonging to this group are those having a partially filled d or f subshell. It usually refers to the d-block transition elements of the periodic table.
3 0
3 years ago
A car drives with constant speed of 112km/h. How long will it take to travel to distance of 56 kilometers?
SSSSS [86.1K]

112 km is travelled by car in 60 minutes

1 km is travelled by car in 60/112 minutes

56 km is travelled by car in 60/112*56 = 30 minutes  = 0.5 hours

Explanation:

3 0
3 years ago
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